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Can one prove that for a prime number $p \geq 2$ and two given integers $d>0$ and $c$, the number of integer solutions $(X,Y)$ to the equation $X^p-dY^p=c$ inside the box $|X| \leq N$, $|Y| \leq N$ is bounded above by $K \log^{p-1} N$, where $K$ is a constant which depends only on $c$ and $d$?

Such a result holds for example when $p=2$ from the theory of Pell's equation, but I could not generalize the arguments there to higher degrees, one reason being that we have more than one fundamental unit in the real extension $\mathbb{Q}(\sqrt[p]{d})$.

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For $p>2$ you get an even tighter bound $K \log^{(p-1)/2} N$. The explanation is because in that case, the group of units in the number field given by the $p$'th root of $d$ has rank $r+s-1 = (p-1)/2$. I can add more details if you want, but the upshot is that if the number of solutions grew faster than this bound, then the ring of algebraic integers in this field would have a larger group of units than allowed by the $r+s-1$ formula, where $r$ is the number of real embeddings (i.e. 1) and $2s$ is the number of complex embeddings (the proof of Dirichlet's unit theorem contains all that you need. In fact, you only need half of the proof, namely the part that shows that $r+s-1$ is an upper bound for the rank of the units).

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    Thank you very much for this. Could you please give a brief explanation for the "growing faster" bit? (I am familiar with algebraic number theory, so an outline is enough) – 2017-02-13
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    Growing faster means that for any fixed $K$, there is an $N$ that violates the bound $K \log^{(p-1)/2} N$. But then, by using the fact that the number of ideals with Norm $c$ is finite, and that the principal ideals form a finite-index subset of that, then you would find "too many" elements generating the same ideal (if two elements generate the same principal ideal, then their quotient is a unit, and if the bound fails, you would find too many units, meaning, more than the $r+s-1$ rank allowed by Dirichlet's theorem). A bit short, but the proof of Dirichlet contains all details needed. – 2017-02-13
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    Actually, come to think of it, the proof of Direchlet that the upper bound $r+s-1$ is sharp uses a similar argument but in the reverse direction. The proof constructs many elements of bounded-norm, and essentially argues that if there were not $r+s-1$ independent units, then you would end up with infinitely many ideals with Norm bounded by a finite number. – 2017-02-13
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    Many thanks. What I was missing when trying the same approach that you suggested is that the units that I keep producing must actually be linearly independent, which, as you mention, would indeed violate the r+s-1 rank. Reducing it to (p-1)/2 is also an easy and nice trick, since we can work with the imaginary field $\mathbb{Q}(\zeta_p\sqrt[p]{d})$ instead of $\mathbb{Q}(\sqrt[p]{d})$ – 2017-02-13
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    Looking at it some more, I think now that for $p>2$ the bound $K \log^{(p-1)/2} N$ is probably still much too high because we're not looking for general algebraic integers $c_0 + c_1 d^{1/p} + \cdots + c_{p-1} d^{(p-1)/p}$ but rather in a subspace $c_2 = \cdots = c_{p-1} = 0$. Since for $p>2$ the number of these linear equations is at least $(p-1)/2$, the number of solutions should be bounded by some $K$ (independent of $N$). – 2017-02-13